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Tuesday, December 8 ∗∗ More on Stokes’ Theorem1. Let F = 〈y2, x2, z2〉. Show that ∫
C1F ·dr =
∫C2
F ·drfor any two closed curves as shown lying on a cylinder about the z-axis.
980 C H A P T E R 17 FUNDAMENTAL THEOREMS OF VECTOR ANALYSIS
16. Let S be the portion of the plane z = x contained in the half-cylinder of radius R depicted in Figure 17. Use Stokes’ Theorem tocalculate the circulation of F = ⟨z, x, y + 2z⟩ around the boundary ofS (a half-ellipse) in the counterclockwise direction when viewed fromabove. Hint: Show that curl(F) is orthogonal to the normal vector tothe plane.
R
−R
FIGURE 17
17. Let I be the flux of F =〈ey, 2xex
2, z2
〉through the upper hemi-
sphere S of the unit sphere.(a) Let G =
〈ey, 2xex
2, 0
〉. Find a vector field A such that
curl(A) = G.(b) Use Stokes’ Theorem to show that the flux of G through S is zero.Hint: Calculate the circulation of A around ∂S .(c) Calculate I . Hint: Use (b) to show that I is equal to the flux of〈0, 0, z2
〉through S .
18. Let F = ⟨0, −z, 1⟩. Let S be the spherical cap x2 + y2 + z2 ≤ 1,where z ≥ 12 . Evaluate
∫∫
SF · dS directly as a surface integral. Then
verify that F = curl(A), where A = (0, x, xz) and evaluate the surfaceintegral again using Stokes’ Theorem.
19. Let A be the vector potential and B the magnetic field of the infinitesolenoid of radius R in Example 4. Use Stokes’ Theorem to compute:(a) The flux of B through a circle in the xy-plane of radius r < R(b) The circulation of A around the boundary C of a surface lyingoutside the solenoid
20. The magnetic field B due to a small current loop (which weplace at the origin) is called a magnetic dipole (Figure 18). Letρ = (x2 + y2 + z2)1/2. For ρ large, B = curl(A), where
A =〈− y
ρ3,
x
ρ3, 0
〉
(a) Let C be a horizontal circle of radius R with center (0, 0, c), wherec is large. Show that A is tangent to C.(b) Use Stokes’ Theorem to calculate the flux of B through C.
Rc
Current loop
A
z
y
x
FIGURE 18
21. Auniform magnetic field B has constant strengthb in the z-direction[i.e., B = ⟨0, 0, b⟩].
(a) Verify that A = 12 B × r is a vector potential for B, where r =⟨x, y, 0⟩.(b) Calculate the flux of B through the rectangle with vertices A, B,C, and D in Figure 19.
22. Let F =〈−x2y, x, 0
〉. Referring to Figure 19, let C be the closed
path ABCD. Use Stokes’ Theorem to evaluate∫
CF · dr in two ways.
First, regard C as the boundary of the rectangle with vertices A, B, C,and D. Then treat C as the boundary of the wedge-shaped box with anopen top.
FIGURE 19
23. Let F =〈y2, 2z + x, 2y2
〉. Use Stokes’ Theorem to find a plane
with equation ax + by + cz = 0 (where a, b, c are not all zero) suchthat
∮
CF · dr = 0 for every closed C lying in the plane. Hint: Choose
a, b, c so that curl(F) lies in the plane.
24. Let F =〈−z2, 2zx, 4y − x2
〉, and let C be a simple closed curve in
the plane x + y + z = 4 that encloses a region of area 16 (Figure 20).Calculate
∮
CF · dr, where C is oriented in the counterclockwise direc-
tion (when viewed from above the plane).
FIGURE 20
25. Let F =〈y2, x2, z2
〉. Show that
∫
C1F · dr =
∫
C2F · dr
for any two closed curves lying on a cylinder whose central axis is thez-axis (Figure 21).
FIGURE 21SOLUTION:Let A be the region of the cylinder bounded by C1 and C2, oriented via the outward pointing
normals. Thus ∂A =C1 −C2. Hence∫C1
F ·dr−∫
C2F ·dr =
∫∂A
F ·dr =∫ ∫
A(curlF) ·dS =
∫ ∫A〈0,0,2(x − y)〉 ·nd A = 0
since n has z-component 0. Thus∫
C1F ·dr = ∫C2 F ·dr
2. Consider the surface T which is the intersection of the plane x+2y+3z = 1 with the first octant.
(a) Draw a picture of T .
SOLUTION:
(b) Use Stokes’ Theorem to evaluate∫∂T
F ·dr for F = 〈y, −2z, 4x〉.SOLUTION:A normal vector v to the plane is 〈1,2,3〉 so a unit normal vector for T is 1p
14〈1,2,3〉.
By Stokes’, we need to evaluate∫ ∫T
(curlF) ·nd A =∫ ∫
T〈2,−4,−1〉 ·nd A =
∫ ∫T
−9p14
d A = −9p14
Area(T ).
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Area(T ) = 12 |a × b| where a = 〈−1,1/2,0〉 and b = 〈0,−1/2,1/3〉, so Area(T ) = 12√
718 . So∫ ∫
T (curlF) ·nd A = −92p14 ·p
73p
2=−34 .
Alternative approach: Parametrize T by r (u, v) = (u, v, 13 (1−u −2v) with domain
D = {0 ≤ u ≤ 1 and 0 ≤ v ≤ 12− u
2}.
ru × rv =
∣∣∣∣∣∣∣i j k1 0 −1/30 1 −2/3
∣∣∣∣∣∣∣= 〈1/3,2/3,1〉So flux = ∫ ∫D〈2,−4,−1〉 · 〈1/3,2/3,1〉d vdu = ∫ ∫D −3d A =−3 Area(D) =−3/4.
3. Carefully explain how Green’s Theorem is actually a special case of Stokes’ Theorem.
SOLUTION:
Green’s theorem is just Stokes’ theorem for a surface and vector field that lie totally inside the
xy-plane. Let’s consider the situation of Green’s theorem — we start with a vector field in the xy-
plane, which looks like 〈P (x, y),Q(x, y)〉, and a region D inside the plane with some boundarycurves ∂D . Let’s think about these living inside of the xy-plane in 3-space, R3, just extending in
the most obvious way, so that we can see what Stokes’ theorem says. Let us now have the vector
field F on R3 defined byF(x, y, z) = 〈P (x, y),Q(x, y),0〉
The region D is now a surface which happens to lie entirely inside the plane z = 0. Computingthe curl of F gives
curl(F ) =
∣∣∣∣∣∣∣i j k∂∂x
∂∂y
∂∂z
P Q 0
∣∣∣∣∣∣∣= 〈−∂Q
∂z,∂P
∂z,∂Q
∂x− ∂P∂y
〉 = 〈0,0, ∂Q∂x
− ∂P∂y
〉
as the functions P and Q only depend on x and y . Looking at the flux of the curl of this vector
field F through the surface with upwards normal n = 〈0,0,1〉, we have the following expression:
Flux through D of curl(F) =Ï
Dcurl(F) ·ndS =
ÏD〈0,0, ∂Q
∂x− ∂P∂y
〉 · 〈0,0,1〉dS =Ï
D
∂Q
∂x− ∂P∂y
dS
Technically, we should call the surface living inside of R3 something like D ′, and parametrizeby r (x, y) = (x, y,0), where the domain for (x, y) is D . In this case, the surface integral abovebecomes exactly a standard double integral in the plane:Ï
D
∂Q
∂x− ∂P∂y
d A
On the other hand, Stokes’ theorem gives us a second expression which computes the flux of
curl(F): the line integral of F along the boundary of D . Note that with upward normal, the rulefor the direction is exactly the same as the one for Green’s theorem – left arm points in walking
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along the direction where your head points up in the direction of the normal. So, we have
exactly the statement of Green’s theorem:
Flux through D of curl(F) =Ï
D
∂Q
∂x− ∂P∂y
d A
and
Flux through D of curl(F) =∫∂D
F ·dr
4. Work the following problem.
980 C H A P T E R 17 FUNDAMENTAL THEOREMS OF VECTOR ANALYSIS
16. Let S be the portion of the plane z = x contained in the half-cylinder of radius R depicted in Figure 17. Use Stokes’ Theorem tocalculate the circulation of F = ⟨z, x, y + 2z⟩ around the boundary ofS (a half-ellipse) in the counterclockwise direction when viewed fromabove. Hint: Show that curl(F) is orthogonal to the normal vector tothe plane.
R
−R
FIGURE 17
17. Let I be the flux of F =〈ey, 2xex
2, z2
〉through the upper hemi-
sphere S of the unit sphere.(a) Let G =
〈ey, 2xex
2, 0
〉. Find a vector field A such that
curl(A) = G.(b) Use Stokes’ Theorem to show that the flux of G through S is zero.Hint: Calculate the circulation of A around ∂S .(c) Calculate I . Hint: Use (b) to show that I is equal to the flux of〈0, 0, z2
〉through S.
18. Let F = ⟨0, −z, 1⟩. Let S be the spherical cap x2 + y2 + z2 ≤ 1,where z ≥ 12 . Evaluate
∫∫
SF · dS directly as a surface integral. Then
verify that F = curl(A), where A = (0, x, xz) and evaluate the surfaceintegral again using Stokes’ Theorem.
19. Let A be the vector potential and B the magnetic field of the infinitesolenoid of radius R in Example 4. Use Stokes’ Theorem to compute:(a) The flux of B through a circle in the xy-plane of radius r < R(b) The circulation of A around the boundary C of a surface lyingoutside the solenoid
20. The magnetic field B due to a small current loop (which weplace at the origin) is called a magnetic dipole (Figure 18). Letρ = (x2 + y2 + z2)1/2. For ρ large, B = curl(A), where
A =〈− y
ρ3,
x
ρ3, 0
〉
(a) Let C be a horizontal circle of radius R with center (0, 0, c), wherec is large. Show that A is tangent to C.(b) Use Stokes’ Theorem to calculate the flux of B through C.
Rc
Current loop
A
z
y
x
FIGURE 18
21. Auniform magnetic field B has constant strengthb in the z-direction[i.e., B = ⟨0, 0, b⟩].
(a) Verify that A = 12 B × r is a vector potential for B, where r =⟨x, y, 0⟩.(b) Calculate the flux of B through the rectangle with vertices A, B,C, and D in Figure 19.
22. Let F =〈−x2y, x, 0
〉. Referring to Figure 19, let C be the closed
path ABCD. Use Stokes’ Theorem to evaluate∫
CF · dr in two ways.
First, regard C as the boundary of the rectangle with vertices A, B, C,and D. Then treat C as the boundary of the wedge-shaped box with anopen top.
FIGURE 19
23. Let F =〈y2, 2z + x, 2y2
〉. Use Stokes’ Theorem to find a plane
with equation ax + by + cz = 0 (where a, b, c are not all zero) suchthat
∮
CF · dr = 0 for every closed C lying in the plane. Hint: Choose
a, b, c so that curl(F) lies in the plane.
24. Let F =〈−z2, 2zx, 4y − x2
〉, and let C be a simple closed curve in
the plane x + y + z = 4 that encloses a region of area 16 (Figure 20).Calculate
∮
CF · dr, where C is oriented in the counterclockwise direc-
tion (when viewed from above the plane).
FIGURE 20
25. Let F =〈y2, x2, z2
〉. Show that
∫
C1F · dr =
∫
C2F · dr
for any two closed curves lying on a cylinder whose central axis is thez-axis (Figure 21).
FIGURE 21SOLUTION:
(a) Parametrize the circle at height c and radius R centered on the z axis by r(t ) = (R cos(t ),R sin(t ),c).Then, the tangent is given by
r′(t ) = (−R sin(t ),R cos(t ),0)
Substituting x(t ) = R cos(t ) and so on into the vector field A, we get what the vector field isat the corresponding point:
A(r(t )) = 〈−R sin(t )ρ3
,R cos(t )
ρ3,0〉
It doesn’t matter what ρ is exactly — we just want to know if the two vectors are multiples
of each other at r(t ), which they are:
r′(t ) · 1ρ3
= A(r(t ))
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(b) Let D be the horizontal disk enclosed by C . We’ll compute the flux with upward normal.
By Stokes’ theorem, ÏD
B ·ndS =Ï
Dcurl(A) ·ndS =
∫C
A ·dr
Here C is traversed counterclockwise as seen from above to match the upward normal.
Then, using our parametrization from before:∫C
A ·dr =∫ 2π
0A(r(t )) · r′(t )d t =
∫ 2π0
r′(t )ρ3
· r′(t )d t =∫ 2π
0
R2
(R2 + c2) 32d t
as r′(t ) · r′(t ) = R2 sin(t )2 +R2 cos(t )2 +02 and ρ3 =(√
(R cos(t ))2 + (R sin(t ))2 + c2)3
Because R and c are fixed, we get a final answer immediately ofÏD
B ·ndS = 2πR2
(R2 + c2) 32